Question

The value of $\int \frac{\sin \theta d\theta }{(4+{\cos }^2\theta )(2-{\sin }^2\theta )}$ is equal to (where $C$ is integration constant)

• A. $\frac{1}{6}{\tan }^{-1}\left(\frac{\cos \theta }{2}\right)-\frac{1}{3}{\tan }^{-1}\left(\cos \theta \right)+C$
• B. $\frac{1}{6}{\tan }^{-1}\left(\frac{\cos \theta }{2}\right)-\frac{1}{3}{\tan }^{-1}\left(\sin \theta \right)+C$
• C. $\frac{1}{3}{\tan }^{-1}\left(\frac{\cos \theta }{2}\right)-\frac{1}{6}{\tan }^{-1}\left(\sin \theta \right)+C$
• D. $\frac{1}{3}{\tan }^{-1}\left(\frac{\cos \theta }{2}\right)-\frac{1}{6}{\tan }^{-1}\left(\cos \theta \right)+C$

Answer 1

The correct option is A $\frac{1}{6}{\tan }^{-1}\left(\frac{\cos \theta }{2}\right)-\frac{1}{3}{\tan }^{-1}\left(\cos \theta \right)+C$

Let $I=\int \frac{\sin \theta d\theta }{(4+{\cos }^2\theta )(2-(1-{\cos }^2\theta ))}$

= $\int \frac{\sin \theta d\theta }{(4+{\cos }^2\theta )(1+{\cos }^2\theta )}$
Put $\cos \theta =t$
$\Rightarrow -\sin \theta d\theta =dt\Rightarrow \sin \theta d\theta =-dt$
$\Rightarrow I=-\int \frac{dt}{(4+t^2)(1+t^2)}$
From Partial Fraction, we can write
$\frac{1}{(4+t^2)(1+t^2)}=\frac{At+B}{4+t^2}+\frac{Ct+D}{1+t^2}$
$1=(At+B)(1+t^2)+(Ct+D)(4+t^2)$
$1=(B+4D)+(A+4C)t+(B+D)t^2+(A+C)t^3$

Put $t=0,$ we have $B+4D=1\cdots \left(i\right)$
Equating the coeff. of $t$ on both sides, we have $A+4C=0\cdots \left(ii\right)$
Equating the coeff. of $t^2$ and $t^3$ respectively
we obtain $B+D=0\cdots \left(iii\right)$ and
$A+C=0\cdots \left(iv\right)$
Solving $\left(i\right),\left(ii\right),\left(iii\right)$ and $\left(iv\right),$ we obtain
$A=C=0,B=-\frac{1}{3}$ and $D=\frac{1}{3}$
$\therefore \frac{1}{(4+t^2)(1+t^2)}=\frac{-\frac{1}{3}}{4+t^2}+\frac{\frac{1}{3}}{1+t^2}$
$I=-\left[\frac{-1}{3}\int \frac{1}{4+t^2}dt+\frac{1}{3}\int \frac{1}{1+t^2}dt\right]$
$=-[\frac{-1}{6}{\tan }^{-1}\frac{t}{2}+\frac{1}{3}{\tan }^{-1}t]+C$
$=\frac{1}{6}{\tan }^{-1}\left(\frac{\cos \theta }{2}\right)-\frac{1}{3}{\tan }^{-1}\left(\cos \theta \right)+C$